Prove that a function is continuous at x =0

Question

I need to prove that $f$ continuous at $(x)=0$ using a $\epsilon$- proof $$ f(x) = \begin{cases} x/(1-x),&x\geq 0 \\ x/(1+x),&x \leq 0 \end{cases} $$

So this is what I have so far:

Let $\epsilon > 0$. Let $\delta <$ (something related to epsilon). Then, $$|f(x)-f(0)| = |f(x)|.$$

I am struggling to figure out what I should set $\delta$ less than in order to prove that $|f(x)| < \epsilon$. Could somebody please help me?

Answer 1

Hint:

If $ \left|x\right|\leq\frac{1}{2}$ then $\left|f\left(x\right)\right|\leq2\left|x\right|$